Question

How do you solve $2\left( {{5}^{6x}} \right)-9\left( {{5}^{4x}} \right)+13\left( {{5}^{2x}} \right)-6=0$?

Answer 1

Hint: We reform the given equation with the help of assumption of ${{5}^{2x}}=a$. Then we use the vanishing method. In this method we find a number $a$ such that for $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We assume $f\left( a \right)=2{{a}^{3}}-9{{a}^{2}}+13a-6$ and take the value of $a$ as 1.

Complete step by step solution:
We have an equation of $2\left( {{5}^{6x}} \right)-9\left( {{5}^{4x}} \right)+13\left( {{5}^{2x}} \right)-6=0$. We assume ${{5}^{2x}}=a$.
We now apply the concept of indices where we have the identity of ${{p}^{mn}}={{\left( {{p}^{m}} \right)}^{n}}$.
Therefore, ${{5}^{6x}}={{\left( {{5}^{2x}} \right)}^{3}},{{5}^{4x}}={{\left( {{5}^{2x}} \right)}^{2}}$ and replacing the value we get ${{5}^{6x}}={{a}^{3}},{{5}^{4x}}={{a}^{2}}$.
The equation becomes $2{{a}^{3}}-9{{a}^{2}}+13a-6=0$. It’s a cubic equation of $a$.
We use vanishing method to solve the problem.
We find the value of $a$ for which the function $f\left( a \right)=2{{a}^{3}}-9{{a}^{2}}+13a-6=0$. We take $a=1$.
We can see $f\left( 1 \right)=2\times {{1}^{3}}-9\times {{1}^{2}}+13\times 1-6=2-9+13-6=0$.
So, the root of the $f\left( a \right)=2{{a}^{3}}-9{{a}^{2}}+13a-6$ will be the function $\left( a-1 \right)$.
Therefore, the term $\left( a-1 \right)$ is a factor of the polynomial $2{{a}^{3}}-9{{a}^{2}}+13a-6$.
We can now divide the polynomial $2{{a}^{3}}-9{{a}^{2}}+13a-6$ by $\left( a-1 \right)$.
\[a-1\overset{2{{a}^{2}}-7a+6}{\overline{\left){\begin{align}
  & 2{{a}^{3}}-9{{a}^{2}}+13a-6 \\
 & \underline{2{{a}^{3}}-2{{a}^{2}}} \\
 & -7{{a}^{2}}+13a \\
 & \underline{-7{{a}^{2}}+7a} \\
 & 6a-6 \\
 & \underline{6a-6} \\
 & 0 \\
\end{align}}\right.}}\]
We first tried to equate the highest power of the dividend with the highest power of the divisor and that’s why we multiplied with $2{{a}^{2}}$. We get \[2{{a}^{3}}-2{{a}^{2}}\]. We subtract it to get \[-7{{a}^{2}}+13a\]. We again equate with the highest power of the remaining terms. We multiply with $-7a$ and subtract to get \[6a-6\]. At the end we had to multiply with 6 to complete the division. The quotient is \[2{{a}^{2}}-7a+6\].
We still can factor \[2{{a}^{2}}-7a+6\].
We apply grouping method where \[2{{a}^{2}}-7a+6=2{{a}^{2}}-3a-4a+6\].
\[\begin{align}
  & 2{{a}^{2}}-3a-4a+6 \\
 & =a\left( 2a-3 \right)-2\left( 2a-3 \right) \\
 & =\left( 2a-3 \right)\left( a-2 \right) \\
\end{align}\]
Therefore, the factorised form of \[2{{a}^{2}}-7a+6\] is \[\left( 2a-3 \right)\left( a-2 \right)\].
The final factorisation is $2{{a}^{3}}-9{{a}^{2}}+13a-6=\left( a-1 \right)\left( 2a-3 \right)\left( a-2 \right)$.
The solutions for $\left( a-1 \right)\left( 2a-3 \right)\left( a-2 \right)=0$ is $a=1,\dfrac{3}{2},2$ which gives ${{5}^{2x}}=1,\dfrac{3}{2},2$.
We take logarithm both sides to get
For ${{5}^{2x}}=1$
$\begin{align}
  & {{\log }_{5}}\left( {{5}^{2x}} \right)={{\log }_{5}}1=0 \\
 & \Rightarrow 2x=0 \\
 & \Rightarrow x=0 \\
\end{align}$
For ${{5}^{2x}}=\dfrac{3}{2}$
$\begin{align}
  & {{\log }_{5}}\left( {{5}^{2x}} \right)={{\log }_{5}}\left( \dfrac{3}{2} \right) \\
 & \Rightarrow 2x={{\log }_{5}}\left( \dfrac{3}{2} \right) \\
 & \Rightarrow x=\dfrac{1}{2}{{\log }_{5}}\left( \dfrac{3}{2} \right) \\
\end{align}$
For ${{5}^{2x}}=2$
$\begin{align}
  & {{\log }_{5}}\left( {{5}^{2x}} \right)={{\log }_{5}}2 \\
 & \Rightarrow 2x={{\log }_{5}}\left( 2 \right) \\
 & \Rightarrow x=\dfrac{1}{2}{{\log }_{5}}\left( 2 \right) \\
\end{align}$
The final solutions are $x=0,\dfrac{1}{2}{{\log }_{5}}\left( \dfrac{3}{2} \right),\dfrac{1}{2}{{\log }_{5}}\left( 2 \right)$.

Note: We find the value of x for which the function $f\left( x \right)=2\left( {{5}^{6x}} \right)-9\left( {{5}^{4x}} \right)+13\left( {{5}^{2x}} \right)-6$. We can see $f\left( 0 \right)=2\left( {{5}^{6x}} \right)-9\left( {{5}^{4x}} \right)+13\left( {{5}^{2x}} \right)-6=2-9+13-6=0$. So, the root of the $f\left( x \right)=2\left( {{5}^{6x}} \right)-9\left( {{5}^{4x}} \right)+13\left( {{5}^{2x}} \right)-6$ will be the function 0.